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every repeating CF will be a quadratic irrational. The CF you wrote is this: $$\alpha = 0 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \ddots}}}} = 0 + \cfrac{1}{1 + \cfrac{1}{2 + \alpha}}$$ now we can just work with that equation: $$\alpha = 1/(1+1/(2+\alpha))$$ $$1/\alpha - 1 = 1/(2+\alpha)$$ $$(1 - \alpha)/\alpha = 1/(2+\alpha)$$ ...
For $n\in\mathbb{N}$, let $q_n=\exp\frac{2\pi\mathrm{i}\tau}{n}$, so $q_n^n=q$. Use formula $(***)$ from that post, but with $q$ replaced with $q^2$, so it reads $$\small\cfrac{1}{1-q^2+\cfrac{(a+bq^2)(aq^2+b)} {1-q^6+\cfrac{q^2(a+bq^4)(aq^4+b)} {1-q^{10}+\cfrac{q^4(a+bq^6)(aq^6+b)}{1-q^{14}+\cdots}}}} = \frac{(-a^2q^6;q^8)_\infty\,(-b^2q^6;q^8)_\infty} ... 2 You want to find out when m is close to 2^{n/2}. If n is even then m = 2^{n/2} is the best you can do, and the next best are 2^{n/2} \pm 1. If n is odd then the best you can do is m = \lceil 2^{n/2} \rceil or m = \lfloor 2^{n/2} \rfloor. Now the interesting question is how close m can get to \sqrt{2} times a power of 2, and this is a ... 0 Five years ago or so I worked on some "tree-type" continued fractions...$$ T = 1+\frac{\displaystyle 1+\frac{\displaystyle 1+\frac{\displaystyle 1+\frac{1+\cdots\;}{2+\cdots\;} }{\displaystyle 2+\frac{2+\cdots\;}{4+\cdots\;} } }{\displaystyle 2+\frac{\displaystyle 2+\frac{2+\cdots\;}{4+\cdots\;} }{\displaystyle ...